In triangle ABC,if A + C = 2B,then frac{a + c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $A + C = 2B$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines,$\cos B = \frac{a^2 +

Vedclass · Accepted Answer

$2\cos \frac{A - C}{2}$

Vedclass · Answer

(A) Given $A + C = 2B$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac}$,which implies $ac = a^2 + c^2 - b^2$,or $b^2 = a^2 - ac + c^2$.Thus,$\sqrt{a^2 - ac + c^2} = b$.The expression becomes $\frac{a + c}{b}$.By the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$,$c = 2R \sin C$,and $b = 2R \sin B$.Substituting these,we get $\frac{2R(\sin A + \sin C)}{2R \sin B} = \frac{\sin A + \sin C}{\sin B}$.Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \frac{A + C}{2} \cos \frac{A - C}{2}$.Since $A + C = 120^{\circ}$,$\frac{A + C}{2} = 60^{\circ}$,so $\sin \frac{A + C}{2} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Also,$\sin B = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Thus,the expression is $\frac{2 \sin 60^{\circ} \cos \frac{A - C}{2}}{\sin 60^{\circ}} = 2 \cos \frac{A - C}{2}$.

In triangle $ABC$,if $A + C = 2B$,then $\frac{a + c}{\sqrt{a^2 - ac + c^2}}$ is equal to

Similar Questions

The lengths of the sides of a triangle are $13$, $14$ and $15$. If $R$ and $r$ respectively denote the circumradius and inradius of that triangle, then $8R + r =$

In $\Delta ABC$,$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = $

The angles of a triangle are in the ratio $1:3:5$. Find the greatest angle. (in $\pi /9$)

In a $\Delta ABC,$ $2ac \sin \left( \frac{A - B + C}{2} \right)$ is equal to

If in a triangle $ABC$,side $a = (\sqrt{3} + 1) \text{ cm}$,$\angle B = 30^\circ$,and $\angle C = 45^\circ$,then the area of the triangle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module